Problem: $h(x)=x^2+5x-4$ What is the average rate of change of $h$ over the interval $[-2,t]$, in terms of $t$, where $t\neq -2$ ? Your answer must be fully expanded and simplified.
Solution: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $h(-2)=-10$. We are interested in the average rate of change of $h(x)=x^2+5x-4$ over the interval $[-2,t]$ : $\begin{aligned} &\phantom{=}\dfrac{h(t)-h(-2)}{(t)-(-2)} \\\\ &=\dfrac{t^2+5t-4-(-10)}{t-(-2)} \\\\ &=\dfrac{t^2+5t+6}{t+2} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{t^2+5t+6}{t+2}&=\dfrac{(t+2)(t+3)}{t+2} \\\\ &=t+3\text{, for }t\neq -2 \end{aligned}$ Since we are given that $t\neq -2$, the average rate of change of the function is $t+3$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.